How does one calculate the chance that an asteroid hits the Earth and what does this chance mean?

Question

I read in this article (in "Independent"):

An asteroid that is projected to come close to Earth later this year has a 0.41 percent chance of hitting the planet, according to Nasa data. The Center for Near-Earth Object Studies (CNEOS), from Nasa’s Jet Propulsion Laboratory, said the celestial object, known as 2018VP1, is predicted to pass near Earth one day before the US presidential election on 2 November.

As usual, I started wondering. How did the people involved at NASA calculated this 0,41% chance of an impact on the earth? What does it even mean to say this? I had the impression that the trajectory of an object thought to come close to earth within a few months could be calculated exactly.
So how did they calculate this uncertainty? Which variables are involved and why they possess uncertainty?

Answer 1

An object discovered recently (2018, I would assume from the name) will have some uncertainty in its position and 3D velocity. That uncertainty will translate into a bigger uncertainty in its position at some time in the future.

If you take a simple example of an object at the origin, measured to be moving along the x-axis. To keep things simple, assume there are no forces acting upon it, but there is an uncertainty in its three velocity components. i.e. it has velocity components $v_x \pm \Delta v_x$, $0 \pm \Delta v_y$ and $0\pm \Delta v_z$.

Now you have a disc, at position $x_0$ along the x-axis, oriented so the flat part is in the $yz$ plane, and with some radius $r$ and you want to work out whether your object will hit it. Obviously if the velocity has its centrally estimated value it will hit it bang on. But after a time $t= x_0/v_x$, if the velocity components in the $y$ and $z$ directions did have a non-zero value, as suggested by their error bars, then the uncertainty in the y and z coordinates at time $t$ would be $\Delta v_y t$ and $\Delta v_z t$. The probability of hitting the disc would be 1 minus the integral of the probability distribution of the y,z position from a radius $r$ out to infinity. You can imagine this like a cone of possible positions that grows with time, and you are calculating what fraction of the cone cross-sectional area is intercepted by the disc.

The task is also complicated for small asteroids because they can be accelerated by non gravitational effects. These include outgassing and mass loss (more important for comets) and the Yarkovsky effect (anisotropic re-radiation of solar flux) and Poynting-Robertson drag (e.g. Broz et al. 2005), leading to rather unpredictable, gradual changes in the orbit.

With this particular object I think it is just that the orbital parameters are rather imprecise. The closest approach is projected to be about 400,000 km, but with a 3-sigma uncertainty of about 4 million km (and the time of arrival is uncertain by a few days). Only a small fraction of that probability distribution ends up with an impact. Since the object is also about 1-m in size, it is unlikely to much damage and would break up in the atmosphere. Because it is so faint, I doubt the parameters will be improved until a few weeks before the fly-by.

Answer 2

Every measurement, and every digital computation, has limited accuracy: it might not be spot on the true value. It will have an "error bound", a range of values within which the true value will (almost) certainly be found. The further a value is from the measured value, the less likely it is to be the true value.

In a sequence of measurements and calculations, the error bound will grow, the uncertainty increase, fewer possibilities can be ruled out and any given value will have less chance of being the true value.

The likelihood of a given outcome being the true value is then expressed as a probability, say 0.41% or 1 in 240.